Chapter 10, Problem 10.104QP

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of pentane has to be written.

Concept Introduction:

Combustion reaction in general is said to be burning in presence of oxygen.  Alkanes undergo combustion to produce carbon dioxide and water along with a large amount of energy being released.  The general reaction scheme can be represented as,

  $\text{Alkane}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Explanation of Solution

Given hydrocarbon is pentane.  Molecular formula of pentane is ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of pentane can be given as,

    ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, five mol of carbon atoms are present.  This can be balanced by adding coefficient 5 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }5{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are twelve mol hydrogen atoms.  Adding coefficient 6 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }5{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are sixteen mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 8 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_{\text{5}}{\text{H}}_{\text{12}}\text{+}8{\text{O}}_{\text{2}}\stackrel{}{\to }5{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(b)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of hexane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is hexane.  Molecular formula of hexane is ${\text{C}}_6{\text{H}}_{\text{14}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of hexane can be given as,

    ${\text{C}}_6{\text{H}}_{\text{14}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, six mol of carbon atoms are present.  This can be balanced by adding coefficient 6 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_6{\text{H}}_{\text{14}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }6{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are fourteen mol hydrogen atoms.  Adding coefficient 7 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_6{\text{H}}_{\text{14}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }6{\text{CO}}_{\text{2}}\text{+}7{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are nineteen mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 9.5 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_6{\text{H}}_{\text{14}}\text{+}9.5{\text{O}}_{\text{2}}\stackrel{}{\to }6{\text{CO}}_{\text{2}}\text{+}7{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Multiplying the above equation by 2 gives whole numbers for all the compounds.  This can be given as,

    ${\text{2C}}_6{\text{H}}_{\text{14}}\text{+}19{\text{O}}_{\text{2}}\stackrel{}{\to }12{\text{CO}}_{\text{2}}\text{+}14{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(c)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of octane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is octane.  Molecular formula of octane is ${\text{C}}_8{\text{H}}_{\text{18}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of octane can be given as,

    ${\text{C}}_8{\text{H}}_{\text{18}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, eight mol of carbon atoms are present.  This can be balanced by adding coefficient 8 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_8{\text{H}}_{\text{18}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }8{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are eighteen mol hydrogen atoms.  Adding coefficient 9 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_8{\text{H}}_{\text{18}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }8{\text{CO}}_{\text{2}}\text{+}9{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are twenty five mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 12.5 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_8{\text{H}}_{\text{18}}\text{+}12.5{\text{O}}_{\text{2}}\stackrel{}{\to }8{\text{CO}}_{\text{2}}\text{+}9{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Multiplying the above equation by 2 gives whole numbers for all the compounds.  This can be given as,

    ${\text{2C}}_8{\text{H}}_{\text{18}}\text{+}25{\text{O}}_{\text{2}}\stackrel{}{\to }16{\text{CO}}_{\text{2}}\text{+}18{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

(d)

Interpretation Introduction

Interpretation:

Balanced equation for the complete combustion of ethane has to be written.

Concept Introduction:

Refer part (a).

Explanation of Solution

Given hydrocarbon is ethane.  Molecular formula of ethane is ${\text{C}}_2{\text{H}}_{\text{6}}$.  Alkanes and cycloalkanes on complete combustion produce carbon dioxide, water, and heat energy.  The chemical equation for the combustion of ethane can be given as,

    ${\text{C}}_2{\text{H}}_{\text{6}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing carbon atom:  In the product side, only one mol of carbon atom is present while in the reactant side, two mol of carbon atoms are present.  This can be balanced by adding coefficient 2 before ${\text{CO}}_{\text{2}}$ in the product side.  The chemical equation obtained after adding coefficient is,

    ${\text{C}}_2{\text{H}}_{\text{6}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing hydrogen atoms:  In the product side, there are two mol of hydrogen atoms, while in the reactant side, there are six mol hydrogen atoms.  Adding coefficient 3 before water in the product side balances out hydrogen atoms on both sides of equation.  The chemical equation obtained is,

    ${\text{C}}_2{\text{H}}_{\text{6}}\text{+}{\text{O}}_{\text{2}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}3{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Balancing oxygen atoms:  In the product side, there are seven mol oxygen atoms, while in the reactant side, there are only two mol oxygen atoms.  Adding coefficient 3.5 before ${\text{O}}_{\text{2}}$ in the reactant side balances out the oxygen atom on both sides of equation.  This step results in balanced chemical equation.  Therefore, the balanced chemical equation can be given as,

    ${\text{C}}_2{\text{H}}_{\text{6}}\text{+}3.5{\text{O}}_{\text{2}}\stackrel{}{\to }2{\text{CO}}_{\text{2}}\text{+}3{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$

Multiplying the above equation by 2 gives whole numbers for all the compounds.  This can be given as,

    ${\text{2C}}_2{\text{H}}_{\text{6}}\text{+}7{\text{O}}_{\text{2}}\stackrel{}{\to }4{\text{CO}}_{\text{2}}\text{+}6{\text{H}}_{\text{2}}\text{O}\text{+}\text{heat}\text{energy}$